Looking at the periodic table, which element is a non-metal?

How many Zn atoms are there in 4.65 moles of Zn?

Nesterboy [21]

So your answer would pretty much be 2.80 x 10^24. The picture is just the explanation and how you would get that answer.

8 0

3 years ago

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r

weqwewe [10]

Answer:

Therefore $\bigtriangledown H^\circ_{rxn}$ = -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by $\bigtriangledown H^\circ_{rxn}$

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s) $\bigtriangledown H^\circ_{rxn}$ =?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1) $\bigtriangledown H^\circ_{rxn}$ = +157KJ

P₄(g)+6Cl₂(g)→ 4PCl₃(g).............(2) $\bigtriangledown H^\circ_{rxn}$ = -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3) $\bigtriangledown H^\circ_{rxn}$ = -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4) $\bigtriangledown H^\circ_{rxn}$ =4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s) $\bigtriangledown H^\circ_{rxn}$ =( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s) $\bigtriangledown H^\circ_{rxn}$ = - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s) $\bigtriangledown H^\circ_{rxn}$ = -1835 KJ

Therefore $\bigtriangledown H^\circ_{rxn}$ = -1835 KJ

5 0

3 years ago

To get a 4 suggest a way to improve the accuracy or precision of the finding<br> .

elixir [45]

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

3 0

4 years ago

When 38^88Sr decays to 34^84Kr, a(n) ______________ is emitted

navik [9.2K]

Answer:

The correct answer is - alpha particle and positron.

Explanation:

In this question, it is given that, 38^88Sr decays to 34^84Kr, which means there is an atomic number decrease by 4, 38 to 34, and atomic mass decreases by 4 as well 88 to 84.

A decrease in the atomic mass is possible only when there is an emission of the alpha particle as an alpha particle is made of 2 protons and 2 neutrons. If an atom emits an alpha particle, there is a change in atomic number as it decreases by two, and its mass number decreases by four.

So after the emission of an alpha particle, the new atom would be

38^88Sr=> 36^84X => 34^84Kr

so there is also two positron emission that leads to decrease in atomic number by one with each emission:

38^88Sr=> 2^4He+ 36^84X => 36^84X + 2(1^0β+) => 34^84Kr

Positron decay is the conversion of a proton into a neutron with the emission of a positron that causes the atomic number is decreased by one, which causes a change in the elemental identity of the daughter isotope.

3 0

3 years ago

lord [1]

Answer:

$\boxed{6.08 \times 10^{23}}$

Explanation:

Data:

I = 2.15 A

t = 8 min 24 s

T = 26.0 °C

V = 65.4 mL

p = 774.2 To

1. Write the equation for the half-reaction

2H₂O ⟶ O₂ + 4H⁺ + 4e⁻

2. Calculate the moles of oxygen

$p = \text{774.2 To} \times \dfrac{\text{1 atm}}{\text{760 To}} = \text{1.0187 atm}$

V = 0.0654 L

T = (26.0 + 273.15) K = 299.15 K

$\begin{array}{rcl}pV& = & nRT\\1.0189 \times 0.0654 & = & n \times 0.08206 \times 299.15\\0.06662 & = & 24.55n\\\\n & = & \dfrac{0.06662}{24.55}\\\\n & = & 2.714 \times 10^{-3}\\\end{array}$

3. Calculate the moles of electrons

$\text{n} = \text{2.714 $\times 10^{-3}$ mol oxygen} \times \dfrac{\text{4 mol electrons}}{\text{ 1 mol ozygen}} = \text{ 0.01086 mol electrons}$

4. Calculate the number of coulombs

t = 8 min 24 s =504 s

Q = It = 504 s × 2.10 C·s⁻¹= 1058 C

5. Calculate the number of electrons

$\text{No. of electrons} = \text{1058 C} \times \dfrac{\text{1 electron}}{1.602 \times 10^{-19}\text{ C}} = 6.607 \times 10^{21}\text{ electrons}$

6. Calculate Avogadro's number

$N_{\text{A}} = \dfrac{6.607 \times 10^{21}}{0.01086} = \mathbf{6.08 \times 10^{23}}\\\\\text{The experimental value of Avogadro's number is } \boxed{\mathbf{6.08 \times 10^{23}}}$

3 0

4 years ago